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Sum the integers, get squares? (Posted on 2007-04-05) Difficulty: 2 of 5
Can the sum of any 2112 consecutive positive integers be a perfect square?

See The Solution Submitted by K Sengupta    
Rating: 2.0000 (3 votes)

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Solution No Subject | Comment 1 of 5

no, it cannot

if n/2*(2*n+2111)=m*m            m is an integer

then  2*n^2+2111*n-2*m^2=0

and 2111+16m^2=k^2              k is an integer


or (k-t )*(k+t )=2111                     t=4*m

and since 2111 is a prime number

there is only one integer solution k=1056 t=1055

but t=4*m     so m cannot be an integer



  Posted by Ady TZIDON on 2007-04-05 09:24:16
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